# Probability of getting at most 3

In tossing four fair dice, what is the probability of tossing, at most one 3?

This is the beginning of the solution I have:

The number of outcomes of tossing four dice is $$6 \times 6 \times 6 \times 6 =6^4$$.

Outcomes of getting at most one 3 = outcomes of getting no 3 $$+$$ outcomes of getting one 3.

The number of outcomes of getting no 3 is $$5 \times 5 \times 5 \times 5 =5^4$$.

The number of outcomes of getting one 3 is $$4 \times 5^3$$.

But I don't understand why it's $$4\times 5^3$$. Can any one assist?

• There are $4$ places you might have gotten the $3$ and $5$ choices for each of the other $3$ places. – lulu Feb 28 '19 at 16:25

1. Choose one tossing which is going to result in a 3. This can be done in $$\binom{4}{1}$$ ways because there are 4 tossings in total.
2. Since the other tossings must not contain a 3, there are 5 numbers available for them. Since there are 3 tossings left, the number of ways is $$5^3$$
Using the multiplicative property you get $$4 \cdot 5^3$$.